The rate equation relates mathematically the rate of reaction to the concentration of the reactants.

Transcription

1 1.9 Rate Equations Rate Equations The rate equation relates mathematically the rate of reaction to the concentration of the reactants. For the following reaction, aa + bb products, the generalised rate equation is: r = k[a] m [B] n r is used as symbol for rate The unit of r is usually mol dm -3 s -1 m, n are called reaction orders Orders are usually integers 0,1,2 0 means the reaction is zero order with respect to that reactant 1 means first order 2 means second order NOTE: the orders have nothing to do with the stoichiometric coefficients in the balanced equation. They are worked out experimentally The square brackets [A] means the concentration of A (unit mol dm -3 ) k is called the rate constant The total order for a reaction is worked out by adding all the individual orders together (m+n) For zero order: the concentration of A has no effect on the rate of reaction r = k[a] 0 = k For first order: the rate of reaction is directly proportional to the concentration of A r = k[a] 1 For second order: the rate of reaction is proportional to the concentration of A squared r = k[a] 2 The rate constant (k) 1. The units of k depend on the overall order of reaction. It must be worked out from the rate equation 2. The value of k is independent of concentration and time. It is constant at a fixed temperature. 3. The value of k refers to a specific temperature and it increases if we increase temperature For a 1 st order overall reaction the unit of k is s -1 For a 2 nd order overall reaction the unit of k is mol -1 dm 3 s -1 For a 3 rd order overall reaction the unit of k is mol -2 dm 6 s -1 Example (first order overall) Rate = k[a][b] 0 m = 1 and n = 0 - reaction is first order in A and zero order in B - overall order = = 1 - usually written: Rate = k[a] Calculating units of k Remember: the values of the reaction orders must be determined from experiment; they cannot be found by looking at the balanced reaction equation 1. Rearrange rate equation to give k as subject k = Rate [A] 2. Insert units and cancel k = mol dm -3 s -1 mol dm -3 Unit of k = s -1 1

2 Example: Write rate equation for reaction between A and B where A is 1 st order and B is 2 nd order. r = k[a][b] 2 overall order is 3 Calculate the unit of k 1. Rearrange rate equation to give k as subject 2. Insert units and cancel 3. Simplify fraction k = Rate [A][B] 2 k = mol dm -3 s -1 mol dm -3.(moldm -3 ) 2 k = s -1 mol 2 dm -6 Unit of k = mol -2 dm 6 s -1 Continuous Monitoring When we follow one experiment over time recording the change in concentration we call it a continuous rate method. The gradient represents the rate of reaction. The reaction is fastest at the start where the gradient is steepest. The rate drops as the reactants start to get used up and their concentration drops. The graph will eventual become horizontal and the gradient becomes zero which represents the reaction having stopped. concentration Measurement of the change in volume of a gas This works if there is a change in the number of moles of gas in the reaction. Using a gas syringe is a common way of following this. It works quite well for measuring continuous rate but a typical gas syringe only measures 100ml of gas so you don t want at reaction to produce more than this volume. Quantities of reactants need to be calculated carefully. time gas syringe Mg + Cl MgCl The initial rate is the rate at the start of the reaction, where it is fastest. It is obtained by taking the gradient of a continuous monitoring conc vs time graph at time = zero. A measure of initial rate is preferable as we know the concentrations at the start of the reaction Typical Method Measure 50 cm 3 of the 1.0 mol dm 3 hydrochloric acid and add to conical flask. Set up the gas syringe in the stand Weigh 0.20 g of magnesium. Add the magnesium ribbon to the conical flask, place the bung firmly into the top of the flask and start the timer. Record the volume of hydrogen gas collected every 15 seconds for 3 minutes. Large Excess of reactants In reactions where there are several reactants, if the concentration of one of the reactant is kept in a large excess then that reactant will appear not to affect rate and will be pseudo-zero order. This is because its concentration stays virtually constant and does not affect rate. 2

3 Comparing continuous rate curves The higher the concentration/ temperature/ surface area the faster the rate (steeper the gradient) If the acid is in excess of the magnesium or marble chips then the final volume of gas produced will be proportional to the amount of moles of acid Different volumes of the same initial concentrations will have the same initial rate (if other conditions are the same) but will end at different amounts Initial rate method Amount of product e.g. Volume of gas Time (secs) Need to calculate/ compare initial moles of reactants to distinguish between different finishing volumes. e.g. the amount of product is proportional to the moles of reactant The initial rate can be calculated from taking the gradient of a continuous monitoring conc vs time graph at time = zero Initial rate can also be calculated from clock reactions where the time taken to reach a fixed concentration is measured. A Common Clock Reaction (no need to learn details) ydrogen peroxide reacts with iodide ions to form iodine. The thiosulfate ion then immediately reacts with iodine formed in the second reaction as shown below. In clock reactions there are often two successive reactions and an end point is achieved when one limited reactant runs out, resulting in a sudden colour change 2 O 2 (aq) (aq) + 2I (aq) I 2 (aq) O(l) 2S 2 O 3 2 (aq) + I 2 (aq) 2I (aq) + S 4 O 6 2 (aq) When the I 2 produced has reacted with all of the limited amount of thiosulfate ions present, excess I 2 remains in solution. Reaction with the starch then suddenly forms a dark blue-black colour. A series of experiments is carried out, in which the concentration of iodide ions is varied, while keeping the concentrations of all of the other reagents the same. In each experiment the time taken (t) for the reaction mixture to turn blue is measured. By repeating the experiment several times, varying the concentration of a reactant e.g. I, ( keeping the other reactants at constant concentration )you can determine the order of reaction with respect to that reactant The initial rate of the reaction can be represented as (1/t ) Experiment Sulfuric acid ( + ) Starch Water Potassium iodide(i - ) Sodium Thiosulfate S 2 O

4 Working out orders from experimental initial rate data Normally to work out the rate equation we do a series of experiments where the initial concentrations of reactants are changed (one at a time) and measure the initial rate each time. Working out rate order graphically In an experiment where the concentration of one of the reagents is changed and the reaction rate measured it is possible to calculate the order graphically Taking rate equation Rate = k [Y] n Log both sides of equation Log rate = log k + n log [Y] Y = c + m x A graph of log rate vs log [Y] will yield a straight line where the gradient is equal to the order n log (rate) y intercept = log K gradient = n = change in y change in x In this experiment high concentrations with quick times will have the biggest percentage errors. This initial rate data can also be presented in a table. Example: work out the rate equation for the following reaction, A+ B+ 2C D + 2E, using the initial rate data in the table Experiment [A] mol dm - 3 [B] mol dm -3 [C] mol dm Rate mol dm -3 s -1 In order to calculate the order for a particular reactant it is easiest to compare two experiments where only that reactant is being changed If conc is doubled and rate stays the same: order= 0 If conc is doubled and rate doubles: order= 1 If conc is doubled and rate quadruples : order= 2 For reactant A compare between experiments 1 and 2 For reactant A as the concentration doubles (B and C staying constant) so does the rate. Therefore the order with respect to reactant A is first order For reactant B compare between experiments 1 and 3 : As the concentration of B doubles (A and C staying constant) the rate quadruples. Therefore the order with respect to B is 2 nd order For reactant C compare between experiments 1 and 4 : As the concentration of C doubles (A and B staying constant) the rate stays the same. Therefore the order with respect to C is zero order The overall rate equation is r = k [A] [B] 2 The reaction is 3 rd order overall and the unit of the rate constant =mol -2 dm 6 s -1 4

5 Working out orders when two reactant concentrations are changed simultaneously In most questions it is possible to compare between two experiments where only one reactant has its initial concentration changed. If, however, both reactants are changed then the effect of both individual changes on concentration are multiplied together to give the effect on rate. In a reaction where the rate equation is r = k [A] [B] 2 If the [A] is x2 that rate would x2 If the [B] is x3 that rate would x3 2 = x9 If these changes happened at the same time then the rate would x2x9= x 18 Example work out the rate equation for the reaction, between X and Y, using the initial rate data in the table Experiment Initial concentration of Initial concentration X/ mol dm 3 of Y/ mol dm 3 Initial rate/ mol dm 3 s x x x 10 6 For reactant X compare between experiments 1 and 2 For reactant X as the concentration doubles (Y staying constant) so does the rate. Therefore the order with respect to reactant X is first order Comparing between experiments 2 and 3 : Both X and Y double and the rate goes up by 8 We know X is first order so that will have doubled rate The effect of Y, therefore, on rate is to have quadrupled it. Y must be second order The overall rate equation is r = k [X] [Y] 2 The reaction is 3 rd order overall and the unit of the rate constant =mol -2 dm 6 s -1 Calculating a value for k using initial rate data Using the above example, choose any one of the experiments and put the values into the rate equation that has been rearranged to give k. Using experiment 3: r = k [X] [Y] 2 k = r [X] [Y] 2 k = 2.40 x x k = 3.0 x 10-4 mol -2 dm 6 s -1 Remember k is the same for all experiments done at the same temperature. Increasing the temperature increases the value of the rate constant k zero order: Calculating k from Concentration-time graphs For zero order reactants, the rate stays constant as the reactant is used up. This means the concentration of that reactant has no effect on rate. Rate = k [A] 0 so rate = k As the rate is the gradient of the graph on the right, the gradient is also the value of the rate constant. [A] The gradient of this graph is equal to the rate constant Time (min) 5

6 Effect of Temperature on Rate Constant: The Arrhenius Equation Increasing the temperature increases the value of the rate constant k Increasing temperature increases the rate constant k. The relationship is given by the Arrhenius equation k = Ae -Ea/RT where A is a constant R is gas constant and Ea is activation energy. k temperature The Arrhenius equation can be rearranged ln k = constant Ea/(RT) k is proportional to the rate of reaction so ln k can be replaced by ln(rate) From plotting a graph of ln(rate) or ln k against 1/T the activation energy can be calculated from measuring the gradient of the line ln (Rate) 1/T Gradient = - Ea/ R Ea = - gradient x R Example Temperature T (K) 1/T time t (s) 1/t Ln (1/t) ln (Rate) x 1,y /T gradient = y 2 -y 1 x 2 -x 1 The gradient should always be -ve x 2,y 2 In above example gradient =-5680 Ea = - gradient x R (8.31) = x8.31 = J mol -1 use a line of best fit use all graph paper choose points far apart on the graph to calculate the gradient The unit of Ea using this equation will be J mol -1. Convert into kj mol -1 by dividing 1000 Ea = kj mol -1 6

7 Rate Equations and Mechanisms A mechanism is a series of steps through which the reaction progresses, often forming intermediate compounds. If all the steps are added together they will add up to the overall equation for the reaction Each step can have a different rate of reaction. The slowest step will control the overall rate of reaction. The slowest step is called the rate-determining step. The molecularity (number of moles of each substance) of the molecules in the slowest step will be the same as the order of reaction for each substance. e.g. 0 moles of A in slow step would mean A is zero order. 1 mole of A in the slow step would mean A is first order Example 1 overall reaction A + 2B + C D + E Mechanism Step 1 A + B X + D slow Step 2 X + C Y fast Step 3 Y + B E fast r = k [A] 1 [B] 1 [C] o C is zero order as it appears in the mechanism in a fast step after the slow step Example 3 Overall Reaction NO 2 (g) + CO(g) NO(g) + CO 2 (g) Mechanism: Step 1 NO 2 + NO 2 NO + NO 3 Step 2 NO 3 + CO NO 2 + CO 2 NO 3 is a reaction intermediate slow fast NO 2 appears twice in the slow steps so it is second order. CO does not appear in the slow step so is zero order. r = k [NO 2 ] 2 Example 2 r = k [X]1 [C]1 Example 5: S N 1 or S N 2? You don t need to remember the details here. overall reaction A + 2B + C D + E Mechanism Step 1 A + B X + D fast Step 2 X + C Y slow Step 3 Y + B E fast The intermediate X is not one of the reactants so must be replaced with the substances that make up the intermediate in a previous step A + B X + D r = k[a] 1 [B] 1 [C] 1 Example 4 Using the rate equation rate = k[no] 2 [ 2 ] and the overall equation 2NO(g) (g) N 2 (g) O(g), the following three-step mechanism for the reaction was suggested. X and Y are intermediate species. Step 1 NO + NO X Step 2 X + 2 Y Step 3 Y + 2 N O Which one of the three steps is the rate-determining step? Step 2 as 2 appears in rate equation and combination of step 1 and 2 is the ratio that appears in the rate equation. Remember the nucleophilic substitution reaction of haloalkanes and hydroxide ions. This is a one step mechanism 3 - O: C δ + δ - C Br C 3 C 2 Br + O - C 3 C 2 O + Br - slow step The rate equation is r = k [C 3 C 2 Br] [O - ] C 3 C O + :Br - This is called S N 2. Substitution, Nucleophilic, 2 molecules in rate determining step The same reaction can also occur via a different mechanism Overall Reaction (C 3 ) 3 CBr + O (C 3 ) 3 CO + Br Mechanism: (C 3 ) 3 CBr (C 3 ) 3 C + + Br (C 3 ) 3 C + + O (C 3 ) 3 CO The rate equation is r = k [(C 3 ) 3 CBr] slow fast This is called S N 1. Substitution, Nucleophilic, 1 molecule in rate determining step 7